A model train starts from rest at point A 10 seconds after a clock has been star

Question

A model train starts from rest at point A 10 seconds after a clock has been started and speeds up at a constant rate until it reaches point B. It then travels at constant speed from point B to point D and thereafter slows down at a constant rate until it comes to rest at point E. Point C Is halfway between points B and D. Find the time t at which the train passes point C. Explain your reasoning or show your work (i.e. the algebra!). A second train (train 2, not shown) starts from point A 2 seconds after the clock has started and speeds up until it reaches point E at the same time (t = 22 s) as train 1 (the train in part a). Is the average velocity of train 2 greater than, less than or equal to the average velocity of train 1? Explain.

Explanation / Answer

(a) at t = 10 s

u = 0

Applying, d = u t + a t^2 /2

2 = 0 + a (15 - 10)^2 / 2

2 = 25a / 2

a = 0.16 m/s^2

vB = vA + a t = 0 + 0.16(15 - 10)

vB = 0.8 m/s

now it moves with constant speed,

1.6 = vB t

t = 1.6 / 0.8 = 2 sec

tC = 15 + 2 = 17 sec ........Ans

(b) average velocity = displacement / time taken

displacemet is same for both the trains.

but second train takes less time as it was started later.

hence second train have greater average velocity than first train.

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